Help! Derivatives and independent variables in Lagrangians
Erik
L = -i hbar Psi* Psi_t + hbar^2/2m grad(Psi*) grad(Psi) + V Psi* Psi,
where Psi_t is the time-derivative of Psi, grad(Psi) is the gradient
and Psi* is the complex conjugate of Psi. The Euler-Lagrange equations
resulting from this Lagrangian is the Schroedinger equation. The
Euler-Lagrange equations can be obtained by computing the derivates
w.r.t. to the real and imaginary parts of Psi (and its partial
derivates). This gives two equations of motions; one for the real part
of the field and one for the imaginary part of the field.
However, there is a simpler and more convenient way to obtain the
equations of motion: Assert that the field Psi and its complex
conjugate Psi* are "independent" and take partial derivates w.r.t. Psi
and Psi* (and their partial derivates). This gives one equation of
motion for Psi and one for Psi*. I cannot figure out why this works.
Psi and Psi* are simply not independent in any sense I've been able to
imagine.
From what I understand the procedure is justified as simply a
change-of-variables trick ("it works for the real and imaginary parts,
and Psi and Psi* are just different coordinates/variables..."), but
this seems incorrect to me. There is a difference between the change
of variables given by, say,
s = Re(Psi) + Im(Psi)
t = Re(Psi) - Im(Psi)
and the one given by
Psi = Re(Psi) + i Im(Psi)
Psi* = Re(Psi) - i Im(Psi).
The variables s and t can be chosen independently of each other. If
I'm told that s = 17 (at a particular point) there will still be
infinitely many possible values of t (at the particular point).
However, if I'm told that Psi = 1+i I'll automatically know the value
of Psi*. Any change of Psi must be matched by a uniquely determined
change of Psi*. Thus, if Psi and Psi* are "independent" they must be
so in a sense that is much weaker than the sense in which s and t (or
Re(Psi) and Im(Psi)) are independent. Hence my problem with the
justification for treating Psi and Psi* as independent and taking
partial derivatives w.r.t. them.
Why does the trick (always?) work? Why can we find the equation of
motion by taking derivates w.r.t. to Psi instead of treating its real
and imaginary part separately?
Thanks in advance,
Erik
Doug Sweetser
This is something people can argue about passionately, even though only
one side is right :-) I think it has to do with learning about complex
numbers first on the manifold of R^2, then assuming lots of those
properties still hold when using complex number on the manifold C^1.
On the manifold R^2, if you know a complex number (t, x), then you know
that (t, x)* = (t, -x).
Now shift to the manifold C^1. The conjugation operator is an
involutive anti-automorphism that goes from C^1 to C^1. Fancy words!
The translation: involutive indicates the conjugate is its own
inverse, so (z*)* = z, and anti-automorphism indicates (z w)* = w* z*.
There is freedom to choose the basis of complex numbers on C^1. That
is where all the misunderstand arises. People do get introduced to the
polar representation of a complex number, but view that as a little
trick, not as a form of freedom. If a number is sitting in C^1, you do
not have to specify the basis for that number. One can construct an
involutive anti-automorphism which starts in one basis, but ends in
another. The conjugate still works perfectly. Since there are an
infinite number of combinations of representations one could choose,
one does not know what z* is given z on the manifold C^1.
One natural reaction is to say if one starts in one basis, one should
end up in the very same basis. Although reasonable, it goes beyond
anything implied by the definition of an involutive anti-automorphism.
It is an additional constraint. Some people will not concede this
point, which leads to the arguments. Oh well.
doug
quaternions.com
Erik
news:<bqndcq$cir$1...@pcls4.std.com>...
Thanks for the explanation, but not everything is yet clear to me. For
example, the set of complex numbers is not just manifold; it is also a
field (in the algebraic sense) and this fact seems to allow us to do
complex conjugation. Like this:
(a) First, find the additive unit element (aka 0).
(b) Second, find the multiplicative unit element (aka 1).
(c) Third, find the addive inverse of the multiplicative unit element
(aka -1).
(d) Fourth, find two elements such that their squares equal the
additive inverse of the multiplicative unit element (aka +i and -i).
Now that we know 0, 1, -1, +i, -i we can do complex conjugation
without ambiguity (just write the number to be conjugated as a linear
combination (with real coefficients) of 1 and +i, or 1 and -i).
Thus, the physical fields we put in as arguments in the Lagrangian are
fields in the algebraic sense. The latter fact allows us to
unambigiously identify the complex conjugate of any physical field.
Therefore the physical fields and their complex conjugates are not
independent. What, if anything, is wrong with reasoning?
Thanks in advance,
Erik
Doug Sweetser
It sounds to me like you are carefully going about and constructing the
manifold R^2. In the first R, you have the additive inverse, the
multiplicative inverse 1, and its additive inverse, -1. You need
another structure to hold i and its additive inverse -i. Now you are
working in R^2. This is a very natural thing to do :-)
On the manifold of C^1, we need to think of the a complex number as one
number, not two. It is one number that does have two degrees of
freedom. One of those choices in the representation. The only way I
finally accepted this idea was to think of an automorphism from the
standard representation into a polar representation. That certainly
twists around what should happen under complex conjugation.
doug
quaternions.com
Erik
> Hello Erik:
>
> It sounds to me like you are carefully going about and constructing the
> manifold R^2.
Yes, but I'm constructing R^2 from the algebraic field C^1 using
nothing in addition. Because of the structure in C^1 additional to its
manifold structure, we can always determine, without ambiguity, the
complex conjugate of any member of C^1. Or so it seems. Here's how:
Pick any element z of C^1. Set 0 := z - z (this is the additive unit
element of C^1). Now pick any element w of C^1\{0} and set 1 := w/w
(this is the multiplicative unit element of C^1). We can now locate a
set in C^1 that is isomorphic to the non-negative integers (e.g. 2 :=
1+1, 3 := 2+1, etc.), and using this set we can construct a set M in
C^1 that is isomorphic to the non-negative rational numbers. We can
define a total ordering relation on M by "x > y iff x - y belongs to
M". Using M and this total ordering relation we can find the subset P
of C^1 that is isomorphic to the non-negative real numbers by the
usual construction of Dedekind cuts. Denote the union of N and all
additive inverses of elements of P by R. R is a subset of C^1 that is
isomorphic to the real numbers.
Using P and R we can now determine the complex conjugate of any z in
C^1 without ambiguity. z* is the unique number in C^1 satisfying
(i) z* z belongs to P,
(ii) z + z* belongs to R,
(iii) (z - z*)^2 belongs to R\(P\{0}).
Have I not just proved that we have no freedom whatsoever when we
"choose" the complex conjugate of an element in C^1?
> In the first R, you have the additive inverse, the
> multiplicative inverse 1, and its additive inverse, -1. You need
> another structure to hold i and its additive inverse -i. Now you are
> working in R^2. This is a very natural thing to do :-)
>
> On the manifold of C^1, we need to think of the a complex number as one
> number, not two. It is one number that does have two degrees of
> freedom. One of those choices in the representation. The only way I
> finally accepted this idea was to think of an automorphism from the
> standard representation into a polar representation. That certainly
> twists around what should happen under complex conjugation.
Maybe I'm just slow, but I'm afraid I still don't understand. In what
way does it twist around what should happen under complex conjugation?
Complex conjugation seems like a perfectly representation-independent
operation. If two elements of the manifold and algebraic field C^1 are
each others' complex conjugates they will remain so regardless of
which coordinates we choose to work with in the manifold C^1.
Thanks in advance,
Erik
Aaron Denney
> Thanks for the explanation, but not everything is yet clear to me. For
> example, the set of complex numbers is not just manifold; it is also a
> field (in the algebraic sense) and this fact seems to allow us to do
> complex conjugation. Like this:
>
> (a) First, find the additive unit element (aka 0).
> (b) Second, find the multiplicative unit element (aka 1).
> (c) Third, find the addive inverse of the multiplicative unit element
> (aka -1).
> (d) Fourth, find two elements such that their squares equal the
> additive inverse of the multiplicative unit element (aka +i and -i).
>
> Now that we know 0, 1, -1, +i, -i we can do complex conjugation
> without ambiguity (just write the number to be conjugated as a linear
> combination (with real coefficients) of 1 and +i, or 1 and -i).
But at this point we *don't* know +i, and -i. We know one of the
elements we picked must be one, and one must be the other, but we don't
know which is which.
--
Aaron Denney
-><-
Erik
But do we need to know? We can take any point z in C^1 and write it as
z = a + jb
where a,b belong to the subset of C^1 that is isomorphic to the real
numbers, and j is one of the two points in C^1 with the property j^2 =
-1. Conjugation can be done achieved through the operation a+jb -->
a+j(-b) regardless of whether j=+i or j=-i.
Perhaps the true justification for taking partial derivates w.r.t. to
complex fields in the Lagrangian lies elsewhere than in a notion of
"independence" Psi and Psi*. Surely it is possible to prove the
equivalence of the Euler-Lagrange equation in terms of complex fields
and the Euler-Lagrange equations in terms of real- and imaginary
parts?
Regards,
Erik
Doug Sweetser
z and z* are definitely independent variables in C^1.
Look at it this way. Complex numbers have two degrees of freedom.
When working in R^2 as you were introduced to the topic, there are two
real numbers to play with. In C^1, that is no longer the case. z is
one number, and one number only. On the manifold C^1, you will want to
do any sort of function that you can do in R^2. You need two ways to
manipulate complex numbers so anything possible on R^2 is also possible
on C^1. That is the coverage you get from working with z and z*. The
manifold C^1 is the land of automorphisms, not the land of separate
real and imaginary parts. That is why an identity automorphism f:z ->
z and an involutive anti-automorphism g:z -> z* can be used to express
any conceivable complex function, no exceptions, without ever breaking
the complex number into two parts, which you so want to do.
A few years ago I tried and failed to convince someone on the net about
this same issue. One thing to do is get out a book on complex analysis
and see those professional mathematicians using z and z* as if they
were independent variables. Seeing published works by professionals
should give you the inspiration to change, but I appreciate it is
difficult. I finally accept all the implications of the 1 in C^1.
doug
quaternions.com
Alfred Einstead
> Consider the Lagrangian
> L = -i hbar Psi* Psi_t + hbar^2/2m grad(Psi*) grad(Psi) + V Psi* Psi,
> where Psi_t is the time-derivative of Psi, grad(Psi) is the gradient
> and Psi* is the complex conjugate of Psi.
>
> equations of motion: Assert that the field Psi and its complex
> works. Psi and Psi* are simply not independent in any sense I've been
> able to imagine.
They are, because...
> There is a difference between the change of variables given by, say,
> s = Re(Psi) + Im(Psi)
> t = Re(Psi) - Im(Psi)
> and the one given by
> Psi = Re(Psi) + i Im(Psi)
> Psi* = Re(Psi) - i Im(Psi).
... there is not a difference, because...
> The variables s and t can be chosen independently of each other.
... which implies therefore that Psi and Psi* can likewise be
chosen independently of each other; namely: Psi = s + it, and
Psi* = s - it.
> If I'm told that s = 17 (at a particular point) there will still be
> infinitely many possible values of t (at the particular point).
> However, if I'm told that Psi = 1+i I'll automatically know the value
> of Psi*.
... the relation Psi* = conjugate of Psi is implied by the equations
of motion, not by any prior constraint. It only comes out after
the equations of motion. Psi and Psi* are also independent before
the equations of motion.
And s and t are also not independent after the equations of motion.
If you go by what things look like after the equations of motion,
the only thing left that's independent is |Psi(x,0)|, arg(Psi(x,0)),
|d/dx Psi(x,0)|, and arg |d/dx Psi(x,0)|. Psi isn't even
independent of itself anymore, then, much less independent of Psi*.
Danny Ross Lunsford
This isn't true. When treating z and z* and independent in physics, there is
almost certainly an implicit assumption about working in R^2 - in physics
this is always related to getting real results when complex numbers are at
hand.
The simplest example is 1/z, which can be written z*/|z|^2. C as a complex
manifold is 1-dimensional, although as a topological manifold it is
2-dimensional. There is one independent variable. z* is itself a
(non-analytic) function of z: z* = 1/z * |z|^2.
What you are thinking of is called an "almost complex manifold", see
Goldberg, "Curvature and Homology".
--
-drl
Doug Sweetser
Your line of logic leaves me confused. What I am trying to say is that
we are working explicity with the manifold C^1, and explicitly not
using R^2, even if this is not common practice. In that way z and z*
are independent variables on C^1. I clearly defined the situation, and
you did not follow the plan.
> The simplest example is 1/z, which can be written z*/|z|^2. C as a
> complex manifold is 1-dimensional, although as a topological manifold
> it is 2-dimensional. There is one independent variable. z* is itself a
> (non-analytic) function of z: z* = 1/z * |z|^2.
I don't see why this is relevant. The function z*^2 is an analytic
function of z* on the C^1 manifold. Lots of functions are
non-analytic. To me, that issue is separate from whether z and z* can
be viewed as completely independent variables on C^1.
doug
quaternions.com
Urs Schreiber
news:d018ba78.0312...@posting.google.com...
> Consider the Lagrangian
>
> L = -i hbar Psi* Psi_t + hbar^2/2m grad(Psi*) grad(Psi) + V Psi* Psi,
[...]
> However, there is a simpler and more convenient way to obtain the
> equations of motion: Assert that the field Psi and its complex
> conjugate Psi* are "independent" and take partial derivates w.r.t. Psi
> and Psi* (and their partial derivates). This gives one equation of
> motion for Psi and one for Psi*. I cannot figure out why this works.
Just add the variation/partial derivative wrt the real part of Psi to +-i
times the variation/partial derivative wrt to the imaginary part of Psi.
Doug Sweetser
This is a curious goal:
> Yes, but I'm constructing R^2 from the algebraic field C^1 using
> nothing in addition.
If we agree that you are analyzing phi and phi* on C^1, you must stay
in C^1 and not think about R^2. This is what is so darn difficult
about understanding manifolds: you must stay in the room. Perhaps this
will relieve some of the pressure: if you work with phi and phi* on the
manifold R^2, then phi* is not independent on phi. Phi* is independent
of phi on C^1.
This gets me nervous:
> We can define a total ordering relation on M by "x > y iff x - y
> belongs to M".
One of the deep ideas about complex numbers is that the set is not a
totally ordered set. This is a partial ordering, because it deals only
with one of the two degrees of freedom (by the way, that was another
way of me finally getting this subtle stuff about complex numbers, by
thinking about degrees of freedom without thinking about numbers like 1
and 3. Math can get so abstract that thinking about numbers doesn't
work so well anymore.)
> Using M and this total ordering relation we can find the subset P
> of C^1 that is isomorphic to the non-negative real numbers by the
> usual construction of Dedekind cuts. Denote the union of N and all
> additive inverses of elements of P by R. R is a subset of C^1 that is
> isomorphic to the real numbers.
To my ear, this sounds like you have constrained complex numbers so
much that they are identical to the real numbers. The real numbers are
a subfield of the field of complex numbers, no doubt about that. Yet
for a more general approach, the total ordering must break down.
>> On the manifold of C^1, we need to think of the a complex number as
>> one number, not two. It is one number that does have two degrees
>> of freedom. One of those choices in the representation. The only
>> way I finally accepted this idea was to think of an automorphism
>> from the standard representation into a polar representation. That
>> certainly twists around what should happen under complex
>> conjugation.
> Maybe I'm just slow, but I'm afraid I still don't understand. In what
> way does it twist around what should happen under complex conjugation?
> Complex conjugation seems like a perfectly representation-independent
> operation. If two elements of the manifold and algebraic field C^1 are
> each others' complex conjugates they will remain so regardless of
> which coordinates we choose to work with in the manifold C^1.
For the record, I don't think you're slow. I recall where I was when
told this property of complex numbers - traveling with my friend Prof.
Guido Sandri to a conference on quaternions in Rome. I thought for
sure he was dead, 100% wrongo. All my experience with complex numbers
was on the manifold R^2. Even when I "claimed" to be working in the
manifold of C^1, my mind still thought in terms of the R^2 manifold
with the two totally ordered sets of real numbers.
Consider a function f which takes a number z and maps it to z'. This
is an example of working in C^1. Think about the kinds of things you
cannot know about z and z'. Somehow we are being more general working
in C^1, but need to be specific about what the generality is. One
think we cannot presume is the basis of z, and _independently_ the
basis of z'. The automorphism f will do its task if z happens to be in
the standard basis and z' is in the polar basis, or visa versa. I
think you may be presuming that people should use the same arbitrary
basis for z and z'. The point is that there is no reason that kind of
consistency between the domain and the range must be maintained.
doug
quaternions.com
Arvind Rajaraman
erit...@yahoo.se (Erik) wrote in message
news:<d018ba78.0312...@posting.google.com>...
> Consider the Lagrangian
>
> L = -i hbar Psi* Psi_t + hbar^2/2m grad(Psi*) grad(Psi) + V Psi* Psi,
>
> where Psi_t is the time-derivative of Psi, grad(Psi) is the gradient
> and Psi* is the complex conjugate of Psi. The Euler-Lagrange equations
> resulting from this Lagrangian is the Schroedinger equation. The
> Euler-Lagrange equations can be obtained by computing the derivates
> w.r.t. to the real and imaginary parts of Psi (and its partial
> derivates). This gives two equations of motions; one for the real part
> of the field and one for the imaginary part of the field.
>
> However, there is a simpler and more convenient way to obtain the
> equations of motion: Assert that the field Psi and its complex
> conjugate Psi* are "independent" and take partial derivates w.r.t. Psi
> and Psi* (and their partial derivates). This gives one equation of
> motion for Psi and one for Psi*. I cannot figure out why this works.
> Psi and Psi* are simply not independent in any sense I've been able to
> imagine.
>
> From what I understand the procedure is justified as simply a
> change-of-variables trick ("it works for the real and imaginary parts,
> and Psi and Psi* are just different coordinates/variables..."), but
> this seems incorrect to me.
> Why does the trick (always?) work? Why can we find the equation of
> motion by taking derivates w.r.t. to Psi instead of treating its real
> and imaginary part separately?
Call the real part of Psi as x and the imaginary part of Psi as y.
Change variables to
Psi_1 =x+iy Psi_2 =x-iy
and any function L(x,y) can be mapped to L(Psi_1,Psi_2).
Now certainly if x and y are complex, there's no problem. Psi_1
and Psi_2 are then linearly independent, and all the usual rules
for changing variables apply. But if some rule works for all complex
(x,y), it must naturally apply to real (x,y) too. So this means we
can treat Psi, Psi^* as independent and get the correct results.
The only way this can fail is if a function has no analytic
continuation; that is, we cannot make x or y complex. Such functions
are unknown (at least to me) in physics contexts.
Danny Ross Lunsford
> Your line of logic leaves me confused. What I am trying to say is that
> we are working explicity with the manifold C^1, and explicitly not
> using R^2, even if this is not common practice. In that way z and z*
> are independent variables on C^1. I clearly defined the situation, and
> you did not follow the plan.
See S. I. Goldberg, "Curvature and Homology", Chapter V. I didn't dream this
up.
>> The simplest example is 1/z, which can be written z*/|z|^2. C as a
>> complex manifold is 1-dimensional, although as a topological manifold
>> it is 2-dimensional. There is one independent variable. z* is itself a
>> (non-analytic) function of z: z* = 1/z * |z|^2.
>
> I don't see why this is relevant. The function z*^2 is an analytic
> function of z* on the C^1 manifold. Lots of functions are
> non-analytic. To me, that issue is separate from whether z and z* can
> be viewed as completely independent variables on C^1.
They are *not* independent in a complex manifold, and I gave a trivial
example as proof - they are independent in an *almost* complex manifold.
The very act of defining a conjugacy relation between z_i and z*_i is what
turns it into a genuine complex manifold. This situation is similar to the
relation between multivectors and p-forms when a metric is at hand.
--
-drl
Doug Sweetser
What exactly is the difference between and *almost* complex manifold
and a complex manifold?
doug
quaternions.com
Erik
> Call the real part of Psi as x and the imaginary part of Psi as y.
> Change variables to
> Psi_1 =x+iy Psi_2 =x-iy
> and any function L(x,y) can be mapped to L(Psi_1,Psi_2).
>
> Now certainly if x and y are complex, there's no problem. Psi_1
> and Psi_2 are then linearly independent, and all the usual rules
> for changing variables apply. But if some rule works for all complex
> (x,y), it must naturally apply to real (x,y) too. So this means we
> can treat Psi, Psi^* as independent and get the correct results.
Many thanks! I find the above intuitively very helpful (although I'm
not sure it would work as a mathematical proof).
> The only way this can fail is if a function has no analytic
> continuation; that is, we cannot make x or y complex. Such functions
> are unknown (at least to me) in physics contexts.
Would the (hypothetical) Lagrangian of the form
L(f,g,...) = ln(1+f^2) + g^2 + ...,
where f and g are real scalar fields, be an example of how introducing
a complex field w=f+ig and considering instead the Lagrangian
L((w+w*)/2,(w-w*)/2i),...) can fail?
Regards,
Erik
Erik
I think you have misunderstood my point. I can assure that I have no
inordinate fondness for any particular representation of the algebraic
field we call the "complex numbers". My point is that I have shown
mathematically that there is no freedom at all in the choice of the
conjugation operator. Only one operator *:C^1 -> C^1 have the
properties we demand of complex conjugation (and all of these
properties are assumed when we write our Lagrangians).
Think of my previous reply to you not as an attempt to reconstruct the
familiar representation of C^1, but as an attempt to prove that we
have no freedom in the choice of conjugation operator. This means that
once we've uniquely defined an element of C^1 we've also uniquely
defined its conjugate. Which in turn means that z and z* are _not_
"independent" in the _usual_ sense.
Your replies to me are not very helpful, because they lack a precise
definition of the unusual meaning of "independence" that must be
indended when it is said that "z and z* are independent".
Regards,
Erik
Doug Sweetser
Let me try again, but this time with the power of ASCII pictures.
Let's start with a complex number z in the complex plane:
| i
z |
|
----------> R
|
|
|
If we want to now graph z*, there is zero choice about where to place
it, because z* is not independent of z in the complex plane:
| i
z |
|
----------> R
|
z* |
|
You would win the discussion if the complex plane was a faithful
representation of C^1. Instead, it is a faithful reprentation of
complex numbers on the manifold R^2. There are two totally ordered
sets, the real numbers and the imaginary numbers. One just does a
reflection over the real number line, and the job is done.
Now lets try and draw the C^1 manifold the same size:
|----------|
| z |
| |
| |
| |
| |
|----------|
Since conjugation is an automorphism, z* has to appear somewhere in
this manifold. Where z* gets draw now depends entirely on where
someone decides to draw in the real line. You might like to draw it
horizontally like the complex plane example above, and I might like to
make it vertical. Clearly both are acceptable. One is free to draw
the real number line _anywhere_ on the C^1 manifold. That is why z* is
independent of z in the traditional sense: you cannot know where z* is
drawn in C^1 given just z. The manifold C^1 is a bit more abstract
than you would like it to be.
doug
quaternions.com
Erik
> Hello Erik:
>
> Let me try again, but this time with the power of ASCII pictures.
>
> Let's start with a complex number z in the complex plane:
>
> | i
> z |
> |
> ----------> R
> |
> |
> |
>
> If we want to now graph z*, there is zero choice about where to place
> it, because z* is not independent of z in the complex plane:
>
> | i
> z |
> |
> ----------> R
> |
> z* |
> |
>
> You would win the discussion if the complex plane was a faithful
> representation of C^1. Instead, it is a faithful reprentation of
> complex numbers on the manifold R^2. There are two totally ordered
> sets, the real numbers and the imaginary numbers. One just does a
> reflection over the real number line, and the job is done.
The algebraic field <R^2,+,*> with the operations defined by
(a,b)+(c,d) = (a+b,c+d)
(a,b)*(c,d) = (ac-bd,ad+bc)
_is_ isomorphic to the complex numbers. If C^1 is not isomorphic to
<R^2,+,*> then it is not directly relevant to this discussion.
> Now lets try and draw the C^1 manifold the same size:
>
> |----------|
> | z |
> | |
> | |
> | |
> | |
> |----------|
>
> Since conjugation is an automorphism, z* has to appear somewhere in
> this manifold. Where z* gets draw now depends entirely on where
> someone decides to draw in the real line. You might like to draw it
> horizontally like the complex plane example above, and I might like to
> make it vertical. Clearly both are acceptable.
No. Clearly one and only one subset of C^1 is isomorphic to the real
numbers. The algebraic field known as the "complex numbers" contains
an additive unit element (let's call it '0') and a multiplicative unit
element (let's call it '1'). One of infinitely many requirements on
the "choice" of real line is that it must pass through these two
elements. This means that it is not true that both a vertical and a
horizontal line are acceptable "choices", because they have only one
point in common and therefore cannot both contain 0 and 1.
> One is free to draw
> the real number line _anywhere_ on the C^1 manifold. That is why z* is
> independent of z in the traditional sense: you cannot know where z* is
> drawn in C^1 given just z. The manifold C^1 is a bit more abstract
> than you would like it to be.
I think of the complex numbers as an algebraic structure that satisfy
certain axioms (i.e. I do _not_ think of e.g. the rectangular or polar
representation or any other representation).
I'm beginning to suspect that your C^1 is in fact not the complex
numbers, but rather a more general structure that lacks some of the
axioms that algebraists demand of the complex numbers. Can you list
all axioms of C^1? Is C^1 a field or is it just a manifold without any
notion of addition and multiplication?
Regards,
Erik
Doug Sweetser
The manifold C^1 has the algebraic structure of a field:
z - z = 0
z z*/|z|^2 = 1
The complex numbers in C^1 form an Abelian group under addition with
the identity element of 0. Ever element in C^1 has an additive
inverse. C^1 without the additive identity element is an Abelian group
under multiplication with an identity element of 1. Every element in
C^1/{0} has an inverse. C^1 is thus a division algebra.
One needs to be skilled in category theory to say how things are alike,
and how things are different. From my reading of the definitions, it
looks like R^2 and C^1 are isomorphic, but not an automorphism since
the sets are different (one contains pairs of real numbers, then other,
one complex number). Check that statement with a local category person.
>> Now lets try and draw the C^1 manifold the same size:
>>
>> |----------|
>> | z |
>> | |
>> | |
>> | |
>> | |
>> |----------|
>>
>> Since conjugation is an automorphism, z* has to appear somewhere in
>> this manifold. Where z* gets draw now depends entirely on where
>> someone decides to draw in the real line. You might like to draw it
>> horizontally like the complex plane example above, and I might like
>> to
>> make it vertical. Clearly both are acceptable.
>
> No. Clearly one and only one subset of C^1 is isomorphic to the real
> numbers. The algebraic field known as the "complex numbers" contains
> an additive unit element (let's call it '0') and a multiplicative unit
> element (let's call it '1'). One of infinitely many requirements on
> the "choice" of real line is that it must pass through these two
> elements. This means that it is not true that both a vertical and a
> horizontal line are acceptable "choices", because they have only one
> point in common and therefore cannot both contain 0 and 1.
I am not disputing there is one additive inverse and one
multiplicative inverse. I am saying there is no unique way to decide
where in the manifold shown above to draw it. In the complex plain,
zero is the intersection of the real and imaginary lines. Such lines
do not exist in C^1. Sure, all the pure real numbers are there, as
well as the purely imaginary numbers, but they are not _organized_ like
they are in R^2. The C^1 manifold can be a complete mess and still
have the properties of an algebraic field. Nothing about the
definition of the manifold C^1 says it must be neatly organized, with
real numbers increasing this way, and imaginary numbers increasing that
way. All that matters is all those numbers happen to be there. You
are presuming orderly lines can be drawn in C^1 based on experience
with R^2, but I don't think that must be the case. A line could be
drawn between 0 and 1, yet not say anything about the rest of the
members in the manifold.
doug
quaternions.com
Erik
the convention that partial derivatives w.r.t. complex fields are
purely formal operations (i.e. they are _not_ defined as limits). We
define @Psi*/@Psi = 0 and compute @L/@Psi _as if_ Psi and Psi* were
real-valued and independent variable. This takes care of the fact that
a Lagrangian may contain non-analytic functions (like the conjugate of
complex field). And because partial derivatives w.r.t. Psi are not
defined as limits we cannot justify their use as simply a change of
variables.
Now consider a Lagrangian of the form
L(@Psi,Psi) = (Psi*)^m Psi^n + ...,
where m and n are positive integers. The variation of L is given by
dL = d((Psi*)^m Psi^n) + d(...)
Here Psi* and Psi must be varied simultaneously giving
d((Psi*)^m Psi^n) = (Psi*+dPsi*)^m (Psi+dPsi)^n - (Psi*)^m Psi^n)
= n (Psi*)^m Psi^(n-1) dPsi + m (Psi*)^(m-1) Psi^n dPsi*
Thus, the term involving dPsi is precisely (@L/@Psi)dPsi. An analogous
observation can be made for dPsi* and @L/@Psi*. Thus, with the purely
formal interpretation of the partial derivatives, we may write
dL = (@L/@Psi)dPsi + (@L/@Psi*)dPsi* + ...
The details are too messy to write out here, but this can be
generalized to any term of the form
(Psi*)^a Psi^b (@Psi/@x)^c (@Psi*/@x)^d ... (@Psi*/@t)^j
With some partial integration in the action integral, terms of the
form A d(@Psi/@x) can be rewritten to the form -(dA/dx)dPsi. At the
end I actually get the Euler-Lagrange equations! Thus, the trick will
work at least for the Lagrangians that can be written as sums of terms
like the one above.
In conclusion, then, the explanation of the trick lies not in a notion
of "independence" of Psi and Psi*, but in a purely formal definition
of partial derivates w.r.t. Psi and Psi* that has no direct relation
to any limit (unlike normal partial derivates).
Thanks for your efforts,
Erik